Volume of Rotation of y = (12)/(x^2 - 16) around Axis

[dariel]

Hi!

I have done a few of these problems fine, but I've got stuck on this:

"The curve \(\displaystyle y=\frac{12}{x^2-16}\) is rotated around the x axis between x=1 meters and x=3 meters. What is the volume of the solid?

Compare the volumes rotated around the x and y axes, what is their shared volume?"

I know the function is how to find the radius, so the area of a 'thin disk' is pi multiplied by the function squared, and the volume is the integration of this.

If I was just integrating the function I could use partial fractions to get

\(\displaystyle y=-\frac{3}{2}(\frac{1}{x+4})+\frac{3}{2}(\frac{1}{x-4})\)

which I could integrate to ln... etc.

But I have to square it and multiply it by pi before I integrate it?? This seems a lot harder than the rest of the problems on the sheet and I feel I am going about it the wrong way.

If I multiply by pi and square the function first, I get

\(\displaystyle y=\frac{144\pi}{(x^2-16)^2}\)

which must then be integrated. I've gone down both paths and ended up doing crazy things without being close to an answer, and it all seems far beyond what we've covered in class. Is there a simple trick I've missed?

Would love a bit of advice!

Thankyou,

Dar

 

[tkhunny]

Have you considered Shells, rather than Disks?
You may wish to consider separately the perfectly cylindrical piece that does not require the calculus.
Does it matter if you investigate 12/(x^2 - 16) or 12/(16-x^2)?

 

[dariel]

I had heard of the shell method, actually, but certainly haven't covered in class

I watched a few videos and it looked promising, no awkward squaring of awkward equations, but I believe to find the width of one shell I need to find 3 minus x in terms of y, giving

\(\displaystyle width=3+\sqrt{\frac{12}{y}+16}\)

This way we would be integrating

\(\displaystyle \int_0^\frac{12}{7} 2 \pi y (3+\sqrt{\frac{12}{y}+16})\)

which doesn't seem much nicer?

I had a think about your other comments, I see the end result will be the same and I assume that looking at it as \(\displaystyle \frac{12}{16-x^2}\) makes it easier somehow, but I don't see it (edited, typo)

Are you suggesting finding the volume of the cylinder radius 0.75 and width 3 (I say width for consistency with the above), shown in blue, and then calculating the rotation of the curve for the area above the curve as far as y=-0.75 (in green)?